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Topic: Calculating Amounts of substances Moles  (Read 16136 times)

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Online Hunter2

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Re: Calculating Amounts of substances Moles
« Reply #15 on: January 24, 2016, 09:06:37 AM »
I do not like to write so long numbers.

1 kilogram = 1000 g. The same with moles 1 kmol = 1000 mol. But it doesnt matter at the end you had to get back to kilogram or tons.

Offline jamesbrown

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Re: Calculating Amounts of substances Moles
« Reply #16 on: January 24, 2016, 10:19:49 AM »
I am so sorry but I really don't understand any of the things you have been talking about could you please explain how you got 9694,73 kg of chlorine and do it in grams please. Also could you explain how I would do the next step which is ratio's i'm guessing? Also Thanks again for putting up with my stupidity as I know for you this is super easy but for me it is quite hard and complex.
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Online Hunter2

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Re: Calculating Amounts of substances Moles
« Reply #17 on: January 24, 2016, 10:45:53 AM »
We have the equation

2 NaCl => 2 Na + Cl2

what means 2 mole of sodiumchloride will convert to 2 mole of sodium and 1 mole of chlorine gas, which is a two-atomic molecule.

We have 16 t NaCl this must fit to x tons sodium + y tons chlorine.

Molar mass of NaCl is 58,5 g/mol, Na 23 g/mol and Cl2 71 g/mol  Units can be extended to kg/kmol
 
The moles must be equal

So u can write n = m(NaCl)/xM(NaCl) = m(Na)/yM(Na) = m(Cl2)/zM(Cl2)

x,y,z= stoechiometric coefficients. x = 2 , y = 2, z =1

m(NaCl) = 16 t = 16000 kg

n = 16000 kg/x 58,6 kg/kmol = 273,5 kmol/x

m (Na) = n *y M(Na)   m(Na) = 273, 5kmol/x * y* 23 kg/kmol = (y/x) 6290,5 kg, y =x = 2, Result 6290,5 kg

m(Cl2) = n*z(M(Cl2)  m(Cl2) =273,5 kmol/x * z * 71 kg/kmol =(z/x)*  19418,5 kg , x =2 z =1

m(Cl2) `=1/2 *19418,5 =9709,25 kg

6290,5 kg + 9709,25 kg =15999,75 kg ~ 16 t

Offline jamesbrown

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Re: Calculating Amounts of substances Moles
« Reply #18 on: January 24, 2016, 11:41:14 AM »
So the mass of Chlorien is 9709,25kg and the mass of Sodium is 6290,5kg
Thank you it is very complex to follow though so I am not certain about this yet.
Q: Did you hear oxygen went on a date with potassium?
A: It went OK.

Online Hunter2

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Re: Calculating Amounts of substances Moles
« Reply #19 on: January 24, 2016, 12:45:12 PM »
To make it more easier

you can balance

m1/xM1 = m2/yM2

m1,2 mass

M1,2 molar mass

x,y stoechoimetric coeficients


Offline jamesbrown

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Re: Calculating Amounts of substances Moles
« Reply #20 on: January 24, 2016, 01:26:36 PM »
I get the mass and molar mass part and the equation but not quite sure what the x,y stoechoimetric coeficients are to be honest. Is it the ratio or what and could you tell me what it is for question 4 (so x,y)
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Re: Calculating Amounts of substances Moles
« Reply #21 on: January 24, 2016, 02:03:01 PM »
You have C6H12O6 => 2 C2H5OH + 2 CO2

Molar Mass sugar is 180 g/mol , Ethanol 46 g/mol carbondioxide 44 g/mol

Coefficients are 1 for sugar and 2 for carbondioxide and alcohol.

2 kg/1*180 kg/kmol = xkg/2*46 kg/kmol

x = 1,022 kg Alcohole

2 kg /1*180 kg/kmol = ykg/2*44 kg/kmol

y= 0,977 kg

1,022 kg + 0,977 kg = 2kg

Offline mikasaur

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Re: Calculating Amounts of substances Moles
« Reply #22 on: January 25, 2016, 12:44:11 PM »
I get the mass and molar mass part and the equation but not quite sure what the x,y stoechoimetric coeficients are to be honest. Is it the ratio or what and could you tell me what it is for question 4 (so x,y)

For questions like these, it's all about tracking your units. I find that writing out expressions where the units cancel is paramount to getting the question right.

C6H12O6  :rarrow: 2 C2H5OH + 2 CO2

MWsugar = 180 g/mol

MWalcohol = 46 g/mol

[itex]\textrm{2 kg sugar } \times\frac{1000\textrm{ g}}{1\textrm{ kg}}\times\frac{1\textrm{ mol sugar}}{180 \textrm{ g sugar}} \times\frac{2 \textrm{ mol alcohol}}{1 \textrm{ mol sugar}}\times \frac{46 \textrm{ g alcohol}}{1 \textrm{ mol alcohol}}[/itex]

See how in the above expression you can start to cancel out units left to right? If you want to always use grams and moles (rather than kg and kmoles) the first thing you do is convert from kg to g. Then you can convert from grams sugar to moles sugar. Then the (sorta) tricky part of using the stoichiometric coefficients. For every 1 mol of sugar consumed you create 2 moles of alcohol! Then you can convert the moles of alcohol to grams of alcohol, et voilĂ . You should be left with only the units you care about, which in this case is a mass of alcohol (in grams). I've left out the conversion back to kg if that's the unit you want.

I would set up your problems like this. Include the balanced chemical equation like Hunter2 said. Cancel units. Double check that the units that remain make sense.
Or you could, you know, Google it.

Offline jamesbrown

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Re: Calculating Amounts of substances Moles
« Reply #23 on: January 25, 2016, 01:07:22 PM »
So for question 5 would you do this

The equation is 2Al2o3 => 4Al + 3O2

The molar mass for Aluminium Oxide is 204

4*27 = 108

1:0.52

5*0.52 = 2.6

Is this right or not?

Also thanks for helping me and I have no clue at all how to do question 6 if you could please help me do that.
Q: Did you hear oxygen went on a date with potassium?
A: It went OK.

Offline mikasaur

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Re: Calculating Amounts of substances Moles
« Reply #24 on: January 25, 2016, 01:14:59 PM »
I don't know if that's right or not. It's hard for me to tell because you have no units in any of your numbers. The molar mass is 204 what? Why are you multiplying 4 by 27? What is the meaning of the number 108? Where does 0.52 come from?

Use units and set up an expression like I did. Cancel the units. Check your answer.

Hunter2 has already mentioned it, but the general approach for stoichiometry questions like these (where they give you a mass of reactant and the reaction goes to completion) is:

1. Write the balanced chemical equation
2. Convert the mass of reactant to moles of reactant
3. Use the stoichiometric coefficients to relate the moles of reactant to the moles of product
4. Convert the moles of product to mass of product
Or you could, you know, Google it.

Offline jamesbrown

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Re: Calculating Amounts of substances Moles
« Reply #25 on: January 25, 2016, 01:17:59 PM »
Sorry man I have a list of questions on page 1 of the post. There you can do it yourself and tell me If I got it correct. Also I have no clue what to do with question 6. But anyway thanks for helping me.
Q: Did you hear oxygen went on a date with potassium?
A: It went OK.

Offline AWK

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Re: Calculating Amounts of substances Moles
« Reply #26 on: January 25, 2016, 01:31:40 PM »
Molar mass of Al2O3 is 102
54/102=0.529
Final result in tonnes is an accepted value
AWK

Offline jamesbrown

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Re: Calculating Amounts of substances Moles
« Reply #27 on: January 25, 2016, 01:35:18 PM »
So AWK is 2.645 tonnes the right answer
Q: Did you hear oxygen went on a date with potassium?
A: It went OK.

Offline Borek

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Re: Calculating Amounts of substances Moles
« Reply #28 on: January 25, 2016, 01:38:28 PM »
Sorry man I have a list of questions on page 1 of the post. There you can do it yourself and tell me If I got it correct. Also I have no clue what to do with question 6. But anyway thanks for helping me.

Sorry if I will sound a bit rash, but if you want to learn try to listen to what others tell you. mikasaur is perfectly correct when he points out your numbers look quite random without units and are hard to follow - which is not OUR problem. It is YOUR problem because that's what makes you lost.

In most cases we have enough experience to guess what you are doing, so we can try to help - but what makes our work just harder makes your work impossible.
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Offline mikasaur

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Re: Calculating Amounts of substances Moles
« Reply #29 on: January 25, 2016, 01:39:34 PM »
Sorry man I have a list of questions on page 1 of the post. There you can do it yourself and tell me If I got it correct. Also I have no clue what to do with question 6. But anyway thanks for helping me.

I know you had a list of questions. When I ask for the units I'm asking if you know them. I know you're getting 27 from the MW of Al and that you're getting the 4 from the stoichiometric coefficient. But what you need to ask yourself is why one of the first things you need to do is multiply 4 by 27 (I don't think it is).

Use the four steps I outlined below, include units, and question 5 should be pretty simple. Question six is not terribly well-worded. I would assume that they mean 2% by mass.

My step 1 is a little harder here and can be skipped by someone well-versed in these sorts of problems but if I were you I would at least try to think about it. Do you think you could make a balanced chemical equation for the process they're describing? Hint: you may have to represent coal as SX, where X is the "stuff" in coal that isn't sulfur.
Or you could, you know, Google it.

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